Compact Sets

$(\Rightarrow)$ Suppose $K$ is compact, and $(x_n)$ is a sequence in $K$ with no subsequence converging to a point of $K$. Then for every $x \in K$ there is $r_x > 0$ such that the open ball $B(x, r_x)$ contains $x_n$ for only finitely many $n$: if instead some $x$ had $B(x, r)$ containing $x_n$ for infinitely many $n$ for every $r > 0$, we could choose $n_k$ increasing with $d(x_{n_k}, x) < 1/k$, and then $(x_{n_k})$ would converge to $x$.

$(\Rightarrow)$ The balls $\{B(x, r_x)\}_{x \in K}$ cover $K$, so compactness gives a finite subcover $B(x_1, r_{x_1}), \dots, B(x_m, r_{x_m})$. Each ball contains $x_n$ for only finitely many $n$, so their finite union contains only finitely many indices $n$ — a contradiction, since a sequence has infinitely many indices. Hence every sequence in $K$ has a subsequence converging to a point of $K$.

$(\Leftarrow)$ Suppose $K$ is sequentially compact (if $K$ is empty both sides are trivial, so assume $K$ is nonempty). Step 1 (Lebesgue number): for every open cover $\mathcal{U}$ of $K$ there is $\delta > 0$ such that every ball $B(x, \delta)$, $x \in K$, lies inside some member of $\mathcal{U}$. Prove it by contradiction: otherwise, for each $n$ pick $x_n \in K$ with $B(x_n, 1/n)$ inside no member of $\mathcal{U}$. By sequential compactness a subsequence $(x_{n_k})$ converges to $x \in K$. There is $U_0 \in \mathcal{U}$ with $x \in U_0$, and $U_0$ open gives $\rho > 0$ with $B(x, \rho) \subseteq U_0$. For all large $k$, $d(x_{n_k}, x) < \rho/2$ and $1/n_k < \rho/2$, so the triangle inequality chain below gives $B(x_{n_k}, 1/n_k) \subseteq U_0$ — contradicting the choice of $x_{n_k}$.

$(\Leftarrow)$ Step 2 (total boundedness): fix $\varepsilon > 0$. If no finite $\varepsilon$-net existed, we could pick a sequence inductively: pick $x_1 \in K$, and pick $x_{n+1} \in K$ with $d(x_{n+1}, x_j) \ge \varepsilon$ for all $j \le n$ — possible because the $\varepsilon$-balls around the finite set $\{x_1, \dots, x_n\}$ do not cover $K$, so a point outside them can be chosen. The resulting sequence has $d(x_m, x_n) \ge \varepsilon$ for all $m \ne n$, so it has no Cauchy subsequence, hence no convergent subsequence — contradicting sequential compactness. Therefore finitely many $\varepsilon$-balls cover $K$.

$(\Leftarrow)$ Step 3: given an open cover $\mathcal{U}$, take its Lebesgue number $\delta > 0$ from Step 1, and cover $K$ by finitely many $\delta$-balls using Step 2 with $\varepsilon = \delta$. By Step 1 each $\delta$-ball lies inside some member of $\mathcal{U}$, so those finitely many members cover $K$. Hence $\mathcal{U}$ has a finite subcover, and since $\mathcal{U}$ was arbitrary, $K$ is compact.

Put $a_n = 1/n$. Then $a_n \in (0,1]$ for every $n$, and $a_n \to 0$.

A subsequence of a convergent sequence converges to the same limit, so every subsequence $(a_{n_k})$ of $(a_n)$ also converges to $0$.

But $0 \notin (0,1]$, so no subsequence converges to a point of $(0,1]$. Hence $(0,1]$ is not sequentially compact, and not compact.

The open-cover picture shows the same failure: however many intervals we take from the cover — the first $N$ of $U_k = \left(\frac{1}{k}, 2\right)$ ($k = 1, \dots, 8$) — their union is always of the form $\left(\frac{1}{N}, 2\right)$, leaving the gap $\left(0, \frac{1}{N}\right]$ — at $k = 8$ that gap is $(0, 1/8]$. Closing the gap would require the point $0$, which is not in the set.

Use the open-cover definition — this argument needs no assumption that $K$ is metric, so it works for a general compact $K$. Let $\{V_i\}_{i \in I}$ be an arbitrary open cover of $f(K)$.

For each $i$, $U_i = f^{-1}(V_i)$ is open in $K$ by continuity of $f$. And if $x \in K$ then $f(x) \in f(K)$, so $f(x) \in V_i$ for some $i$, i.e. $x \in U_i$ — hence $\{U_i\}$ is an open cover of $K$.

Since $K$ is compact, there is a finite subcover: $K \subseteq U_{i_1} \cup \cdots \cup U_{i_n}$. Then the corresponding $V_{i_1}, \dots, V_{i_n}$ cover $f(K)$: if $y = f(x) \in f(K)$, then $x \in U_{i_j}$ for some $j$, so $y = f(x) \in V_{i_j}$.

Hence every open cover of $f(K)$ has a finite subcover, so $f(K)$ is compact. This is the engine of the Extreme Value Theorem (the evt scene) — for continuous $f$ on $[0,1]$, the set $f([0,1])$ is compact, hence closed and bounded in $\mathbb{R}$, so its supremum and infimum are actually attained.